Answer
(a) $$D_f= \mathbb R -\{1\}$$ $$R_f= \mathbb R - \{0\}$$
(b) $$f(f(x))=\frac{1}{1-\frac{1}{1-x}} \\$$ $$ D_{f \circ f}= \mathbb R - \{1,0\}$$
(c) $$f(f(f(x)))=\frac{1}{1-\frac{1}{1-\frac{1}{1-x}}}$$ $$D_{f \circ f \circ f}= \mathbb R - \{1,0\}$$
(d) The image of the graph of $f(f(f(x)))$ has been added.
No, it is not a line since a line has no gap, but this graph has two gaps at the points $(0,0)$ and $(1,1)$.
Work Step by Step
(a) The domain of a real rational function is all real numbers, $\mathbb R$, except those vanishing its denominator.
So the domain of $f$, $D_f$, is $\mathbb R - \{ \text {roots of } (1-x) \}= \mathbb R -\{1\}.$
To find the range of $f$, let $y$ be any real number which lies in the range of $f$. So there exists some $x \in D_f$ such that $y=f(x)$, that is, $$y=\frac{1}{1-x} \quad \Rightarrow \quad y-xy=1 \quad \Rightarrow \quad x= \frac{y-1}{y}.$$The right hand side of last equation can be defined for all real numbers except ${0}$. Thus, the range of $f$, $R_f$, is $\mathbb R - \{0\}$.
(b)$$f(x)=\frac{1}{1-x} \quad \Rightarrow \quad f(f(x))=\frac{1}{1-\frac{1}{1-x}}$$Since $f(f(x))$ has two fractions, the domain of $f(f(x))$, $D_{f \circ f}$ is all real numbers except those vanishing either of denominators.
Thus,$$D_{f \circ f}= \\ \mathbb R -\left ( \left \{ \text{roots of } \left (1-x \right ) \right \} \bigcup \left \{ \text{roots of } \left (1-\frac{1}{1-x} \right ) \right \} \right ) \\ \Rightarrow D_{f \circ f}= \mathbb R - (\{1\} \cup \{0\})=\mathbb R - \{1,0 \}. $$
For $x \in D_{f \circ f}$, we can write $f(f(x))$ as $$f(f(x))=\frac{1}{\frac{1-x-1}{1-x}}= \frac{x-1}{x}.$$
(c)$$f(x)=\frac{1}{1-x} \quad , \quad f(f(x))=\frac{1}{1-\frac{1}{1-x}} \quad \Rightarrow \quad f(f(f(x)))=\frac{1}{1-\frac{1}{1-\frac{1}{1-x}}} $$Since $f(f(f(x)))$ has three fractions, the domain of $f(f(f(x)))$, $D_{f \circ f \circ f}$ is all real numbers except those vanishing either of denominators.
Thus,$$D_{f \circ f \circ f}= \\ \mathbb R -\left ( \left \{ \text{roots of } \left (1-x \right ) \right \} \bigcup \left \{ \text{roots of } \left (1-\frac{1}{1-x} \right ) \right \} \bigcup \left \{ \text{roots of } \left (1-\frac{1}{1-\frac{1}{1-x}} \right ) \right \} \right ) \\ \Rightarrow D_{f \circ f \circ f}= \mathbb R - ( \{1\} \cup \{0\} \cup \varnothing ) = \mathbb R - \{1,0 \}. $$
For $x \in D_{f \circ f \circ f}$, we can write $f(f(f(x)))$ as $$f(f(f(x)))= \frac {1}{\frac{x-(x-1)}{x}}=x.$$
(d) It has been explained in the answer.
